sum of $-\sum_{d \leq x}\mu(d) g(d) \log(d)$

Question

I have seen this sum in "Asymptotic Sieve for primes" If $g(d)$ is a multiplicative function, then $-\sum_{d \leq x}{\mu(d)g(d)\log(d)}$ = the infinite product $\prod_{p}(1-g(p))(1-\frac{1}{p})^{-1}$, Can any one one explain the intermediate step involved